unit 7 mixed stoichiometry problems how many grams of acetylene (c₂h₂) are produced by adding water to 5.00…
unit 7 mixed stoichiometry problems how many grams of acetylene (c₂h₂) are produced by adding water to 5.00 g cac₂? __cac₂ + __h₂o → __c₂h₂ + __ca(oh)₂ given: grams of cac₂ = 5g __cac₂ + 2 h₂o → __c₂h₂ + __ca(oh)₂ 5gcac₂ multiple - choice question how many railroad tracks do you need to answer this question? 4 it should be a rice table! 3 2
Answer
Explanation:
Step1: Balance the chemical equation
$CaC_{2}+2H_{2}O\rightarrow C_{2}H_{2}+Ca(OH)_{2}$
Step2: Calculate molar - masses
The molar mass of $CaC_{2}$: $M_{CaC_{2}}=40.08 + 2\times12.01=64.1\ g/mol$. The molar mass of $C_{2}H_{2}$: $M_{C_{2}H_{2}}=2\times12.01+2\times1.01 = 26.04\ g/mol$.
Step3: Calculate moles of $CaC_{2}$
$n_{CaC_{2}}=\frac{m_{CaC_{2}}}{M_{CaC_{2}}}=\frac{5.00\ g}{64.1\ g/mol}\approx0.078\ mol$
Step4: Use mole - ratio
From the balanced equation, the mole - ratio of $CaC_{2}$ to $C_{2}H_{2}$ is 1:1. So, $n_{C_{2}H_{2}} = n_{CaC_{2}}=0.078\ mol$
Step5: Calculate mass of $C_{2}H_{2}$
$m_{C_{2}H_{2}}=n_{C_{2}H_{2}}\times M_{C_{2}H_{2}}=0.078\ mol\times26.04\ g/mol\approx2.03\ g$
Answer:
$2.03\ g$